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∠AOB=2∠ACB=2∠C∠AOCB=2(180°-∠AOB)=360°-4∠C∠OCAB=(180°-∠AOCB)/2=2∠C-90°Let C' be the point of intersection of AB and COC.Area of ΔCAOC=1/2×AC×AOC×sin(A+∠OCAB)=1/2×AC×AOC×sin(A+2C-90°)=1/2×AC×AOC×sin(90°+C-B)=1/2×AC×AOC×cos(C-B)AC'/C'B=Area of ΔCAOC/Area of ΔCBOC=1/2×AC×AOC×cos(C-B)/(1/2×BC×AOC×cos(C-A))=AC×cos(C-B)/(BC×cos(C-A))AC'/C'B×BA'/A'C×CB'/B'A=1